Author Topic: Q6 TUT 0501  (Read 2741 times)

Victor Ivrii

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Q6 TUT 0501
« on: November 17, 2018, 03:54:44 PM »
Express the general solution of the given system of equations in terms of real-valued functions:
$$\mathbf{x}' = \begin{pmatrix} -3 &0 &2\\ 1 &-1 &0\\ -2 &-1 &0 \end{pmatrix}\mathbf{x}.$$

Boyu Zheng

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Re: Q6 TUT 0501
« Reply #1 on: November 17, 2018, 04:21:46 PM »
\begin{equation*}
det
\begin{pmatrix}
-3\lambda    &0          &2 \\
1         & -1\lambda    &0 \\
-2         & -1            & -\lambda
\end{pmatrix}
=-\lambda^3-4\lambda^2-7\lambda-6=-(\lambda+2)(\lambda+2\lambda+3)=0
\end{equation*}
$$\lambda=-2,\lambda=\sqrt{2}\qquad i-1,\lambda=-\sqrt{2}\qquad i-1$$

when $\lambda$=-2
\begin{equation*}
\begin{pmatrix}
-1          &0           &2 \\
1         & 1             &0 \\
-2         & -1            & 2
\end{pmatrix}

\begin{pmatrix}
x_1            \\       x_2       \\       x_3
\end{pmatrix}=0
\end{equation*}
$$\text{let } x_3=t,x_1=2t,x_2=-2t, x= \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}t$$

when $\lambda=\sqrt{2}\qquad i-1$

\begin{equation*}
\begin{pmatrix}
1         & \sqrt{2}\qquad i            &0 \\
-2         & -1            & -\sqrt{2}\qquad i+1
\end{pmatrix}
\begin{pmatrix}
x_1    \\ x_2 \\ x_3
\end{pmatrix}
=0
\end{equation*}
$$x=\begin{pmatrix} \frac{2}{3}-\frac{i\sqrt{2}\qquad}{3}\\\frac{-1}{3}-\frac{i\sqrt{2}\qquad}{3} \\1 \end{pmatrix}t$$
\begin{equation*}
x(t)=c_1e^{-2t}
\begin{pmatrix}
2\\-2\\1
\end{pmatrix}
+c_2e^{-t}
\begin{pmatrix}
\frac{2}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
-\frac{1}{3}\cos \sqrt{2}\theta+\frac{\sqrt{2}}{3}\sin \sqrt{2}\theta\\
\cos\sqrt{2}\theta
\end{pmatrix}
+c_3e^{-t}i
\begin{pmatrix}
\frac{2}{3}\sin \sqrt{2}\theta-\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
-\frac{1}{3}\sin \sqrt{2}\theta+\frac{\sqrt{2}}{3}\cos \sqrt{2}\theta\\
\sin\sqrt{2}\theta
\end{pmatrix}
\end{equation*}
« Last Edit: November 19, 2018, 05:05:33 PM by Boyu Zheng »

Ruiling Zhao

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Re: Q6 TUT 0501
« Reply #2 on: November 17, 2018, 05:00:23 PM »
Quiz 6

Xiaoyuan Wang

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Re: Q6 TUT 0501
« Reply #3 on: November 17, 2018, 05:26:02 PM »

Qi Cui

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Re: Q6 TUT 0501
« Reply #4 on: November 17, 2018, 05:38:40 PM »
$$det(A-\lambda I) = \left| \begin {array}{ccc} {-3- \lambda}&0&2\\ 1& {-1- \lambda}&0\\ -2&-1&{- \lambda} \end {array} \right| = 0$$
$${\lambda}^3 + 4 {\lambda}^2+ 7{\lambda}+6 = 0$$
$$By\ long\ devision\ method, we\ get\ (\lambda+2)({\lambda}^2+ 2{\lambda} +3) = 0$$
$$({\lambda}^2+ 2{\lambda} +3) : \lambda = {-2 \pm \sqrt{-8} \over 2} = -1\pm \sqrt{2}i$$
$$\quad\therefore \lambda = -2; {-1\pm \sqrt{2}i}$$
$when\ \lambda = -2:$
$$(A-\lambda I)x = 0$$
$$\left[ \begin {array}{ccc} -1&0&2\\ 1&1&0\\ -2&-1&2 \end {array} \right]x = 0$$
$$By\ row\ operation, we\ get: \left[ \begin {array}{ccc} -2&-1&2\\ -1&0&2\\ -0&0&0 \end {array} \right]\left[ \begin {array}{c} x_1\\ x_2\\ x_3 \end {array} \right]= 0$$
$let x_3 = t:$
$$-2x_1-x_2+2t= 0$$
$$-x_1+2t= 0$$
$we\ have$:
$$span(\left[ \begin {array}{c} 2\\ -2\\ 1 \end {array} \right])$$
$By\ similar\ procedure\ as\ above\ shown,we\ can\ get\ other\ two\ eigenvector\ for \lambda = -1\pm \sqrt{2}i : span(\left[ \begin {array}{c} 2- \sqrt{2}i\\ -1- \sqrt{2}i\\ 1 \end {array} \right], \left[ \begin {array}{c} 2+ \sqrt{2}i\\ -1+ \sqrt{2}i\\ 1 \end {array} \right] )$
$$\quad\therefore x(t) = c_1e^{-2t}\left( \begin {array}{c} 2\\ -2\\ 1 \end {array} \right)+c_2e^{(-1+ \sqrt{2}i)t }\left( \begin {array}{c} 2- \sqrt{2}i\\ -1- \sqrt{2}i\\ 1 \end {array} \right)+c_3e^{(-1- \sqrt{2}i)t}\left( \begin {array}{c} 2+ \sqrt{2}i\\ -1+ \sqrt{2}i\\ 1 \end {array} \right)$$
« Last Edit: November 20, 2018, 07:00:21 PM by Qi Cui »

Cheng Qian

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Re: Q6 TUT 0501
« Reply #5 on: November 18, 2018, 12:19:54 AM »
here is my solution in pdf using Latex